Rút gọn :
\(a,A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ b,B=-1^2+2^2-3^2+4^2-...-99^2+100^2\\ c,C=-1^2+2^2-3^2+4^2-...+\left(-1\right)^n\cdot n^2\\ d,D=3\cdot\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ e,E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\\ g,G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\\ h,H=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3+\left(a+b-c\right)^3\\ i,I=\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(c+b\right)\left(c+a\right)\)
Mọi người ơi, giúp mk vs, đc câu nào hay câu ấy ! Help me!!!!!!!!!!!!!!!!!!
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)
e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=\left(2a+2b\right)^2-2\left(a+b+c\right)\left(a+b-c\right)-2\left(a+b\right)^2\)
\(=4\left(a+b\right)^2-2\left(\left(a+b\right)^2-c^2\right)-2\left(a+b\right)^2\)
\(=4\left(a+b\right)^2-2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2=2c^2\)
g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)
\(=\left(2a+2b\right)^2-2\left(a+b+c+d\right)\left(a+b-c-d\right)+\left(2a-2b\right)^2-2\left(a+c-b-d\right)\left(a+d-b-c\right)\)
\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)
\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)
\(=2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)\(=2\left(\left(2a\right)^2-2\left(a+b\right)\left(a-b\right)\right)+2\left(\left(2c\right)^2-2\left(c+d\right)\left(c-d\right)\right)\)
\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)
\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)
bn đăng nhiều quá nên mk làm câu nào hay câu đó nha
mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .
ta dể dàng có được : \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) và
\(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)
áp dụng cái này cho câu h
h) ta có : \(H=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3+\left(a+b-c\right)^3\)
\(=\left(a+b+c\right)^3+\left(a-b-c\right)^3+\left(b-c-a\right)^3+\left(b-c+a\right)^3\)
\(=\left(2a\right)^3-3\left(a+b+c\right)\left(a-b-c\right)\left(2a\right)+\left(2b-2c\right)^3-3\left(b-c-a\right)\left(b-c+a\right)\left(2b-2c\right)\)
\(=8a^3+\left(2b-2c\right)^3-6a\left(a^2-\left(b+c\right)^2\right)-\left(6b-6c\right)\left(\left(b-c\right)^2-a^2\right)\)
\(=2a^3+2\left(b-c\right)^3+6a^2\left(b-c\right)+6a\left(b+c\right)^2\)
\(=...\) bn làm nốt nha
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